Samacheer Kalvi Class 9 Maths Solutions Chapter 2 Real Numbers Exercise 2.5

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.5

 

Question 1. Write the following in the form of \( 5^n \):
(i) 625
(ii) \( \frac { 1 }{ 5 } \)
(iii) \( \sqrt{5} \)
(iv) \( \sqrt{125} \)
Answer:
(i) We need to express 625 as a power of 5. By repeatedly dividing 625 by 5, we find:
\( 625 = 5 \times 5 \times 5 \times 5 = 5^4 \).
So, \( 625 = 5^4 \).
(ii) To write \( \frac { 1 }{ 5 } \) in the form of \( 5^n \), we use the rule that \( \frac { 1 }{ a } = a^{-1} \).
So, \( \frac { 1 }{ 5 } = 5^{-1} \).
(iii) The square root of a number can be written as that number raised to the power of \( \frac { 1 }{ 2 } \).
Therefore, \( \sqrt{5} = 5^{\frac{1}{2}} \).
(iv) First, express 125 as a power of 5: \( 125 = 5 \times 5 \times 5 = 5^3 \).
Now, \( \sqrt{125} = \sqrt{5^3} \). Using the rule \( \sqrt[m]{a^n} = a^{\frac{n}{m}} \), where for a square root \( m=2 \), we get:
\( \sqrt{5^3} = (5^3)^{\frac{1}{2}} = 5^{3 \times \frac{1}{2}} = 5^{\frac{3}{2}} \).
In simple words: We are changing each number into a power of 5. This means finding how many times 5 is multiplied by itself (or its inverse) to get the original number. For roots, we use fractional powers.

🎯 Exam Tip: Remember the exponent rules for negative powers (\( a^{-n} = \frac{1}{a^n} \)) and fractional powers (\( \sqrt[m]{a^n} = a^{\frac{n}{m}} \)) as they are key to solving such problems.

 

Question 2. Write the following in the form of \( 4^n \):
(i) 16
(ii) 8
(iii) 32
Answer:
(i) To write 16 in the form of \( 4^n \), we know that \( 4 \times 4 = 16 \).
So, \( 16 = 4^2 \).
(ii) To write 8 in the form of \( 4^n \):
\( 8 = 4 \times 2 \)
We know that \( 2 = \sqrt{4} = 4^{\frac{1}{2}} \).
So, \( 8 = 4 \times 4^{\frac{1}{2}} \)
Using the exponent rule \( a^m \times a^n = a^{m+n} \), where \( 4 = 4^1 \):
\( 8 = 4^{1+\frac{1}{2}} = 4^{\frac{2}{2}+\frac{1}{2}} = 4^{\frac{3}{2}} \).
(iii) To write 32 in the form of \( 4^n \):
\( 32 = 4 \times 8 \)
From part (ii), we know \( 8 = 4^{\frac{3}{2}} \).
So, \( 32 = 4 \times 4^{\frac{3}{2}} \)
Again, using \( a^m \times a^n = a^{m+n} \):
\( 32 = 4^{1+\frac{3}{2}} = 4^{\frac{2}{2}+\frac{3}{2}} = 4^{\frac{5}{2}} \).
In simple words: We are changing each number into a power of 4. This might mean using fractional powers if the number is not a perfect power of 4. We find out how many times 4 needs to be multiplied by itself to get the original number.

🎯 Exam Tip: When converting to a new base (like 4), it's often helpful to first factorize the number to see its prime components, and then group them to match the new base's components.

 

Question 3. Find the value of
(i) \( (49)^{\frac{1}{2}} \)
(ii) \( (243)^{\frac{2}{5}} \)
(iii) \( (9)^{\frac{-3}{2}} \)
(iv) \( (\frac{64}{125})^{\frac{-2}{3}} \)
Answer:
(i) We need to find the value of \( (49)^{\frac{1}{2}} \). We know that \( 49 = 7^2 \).
So, \( (49)^{\frac{1}{2}} = (7^2)^{\frac{1}{2}} \).
Using the exponent rule \( (a^m)^n = a^{m \times n} \):
\( (7^2)^{\frac{1}{2}} = 7^{2 \times \frac{1}{2}} = 7^1 = 7 \).
(ii) We need to find the value of \( (243)^{\frac{2}{5}} \). First, we find the prime factorization of 243:

From this, we see that \( 243 = 3 \times 3 \times 3 \times 3 \times 3 = 3^5 \).
So, \( (243)^{\frac{2}{5}} = (3^5)^{\frac{2}{5}} \).
Using the exponent rule \( (a^m)^n = a^{m \times n} \):
\( (3^5)^{\frac{2}{5}} = 3^{5 \times \frac{2}{5}} = 3^2 = 9 \).
(iii) We need to find the value of \( (9)^{\frac{-3}{2}} \). We know that \( 9 = 3^2 \).
So, \( (9)^{\frac{-3}{2}} = (3^2)^{\frac{-3}{2}} \).
Using the exponent rule \( (a^m)^n = a^{m \times n} \):
\( (3^2)^{\frac{-3}{2}} = 3^{2 \times \frac{-3}{2}} = 3^{-3} \).
Using the exponent rule \( a^{-n} = \frac{1}{a^n} \):
\( 3^{-3} = \frac{1}{3^3} = \frac{1}{3 \times 3 \times 3} = \frac{1}{27} \).
(iv) We need to find the value of \( (\frac{64}{125})^{\frac{-2}{3}} \).
First, express 64 and 125 as powers of their prime factors: \( 64 = 4^3 \) and \( 125 = 5^3 \).
So, \( (\frac{64}{125})^{\frac{-2}{3}} = (\frac{4^3}{5^3})^{\frac{-2}{3}} \).
This can be written as \( ((\frac{4}{5})^3)^{\frac{-2}{3}} \).
Using the exponent rule \( (a^m)^n = a^{m \times n} \):
\( ((\frac{4}{5})^3)^{\frac{-2}{3}} = (\frac{4}{5})^{3 \times \frac{-2}{3}} = (\frac{4}{5})^{-2} \).
Using the exponent rule \( (\frac{a}{b})^{-n} = (\frac{b}{a})^n \):
\( (\frac{4}{5})^{-2} = (\frac{5}{4})^2 = \frac{5^2}{4^2} = \frac{25}{16} \).
In simple words: For each problem, we change the base number into a power of a smaller number. Then, we multiply the powers together. If the power is negative, we flip the fraction or take 1 divided by the positive power.

🎯 Exam Tip: Always break down the base number into its prime factors first, especially for larger numbers or those with fractional exponents. This simplifies the calculation significantly.

 

Question 4. Use a fractional index to write:
(i) \( \sqrt{5} \)
(ii) \( \sqrt[2]{7} \)
(iii) \( (\sqrt[3]{49})^{5} \)
(iv) \( (\frac{1}{\sqrt[3]{100}})^{7} \)
Answer:
(i) The square root of a number can be written as that number raised to the power of \( \frac{1}{2} \).
So, \( \sqrt{5} = 5^{\frac{1}{2}} \).
(ii) The notation \( \sqrt[2]{7} \) is another way to write the square root of 7. The index 2 indicates a square root.
So, \( \sqrt[2]{7} = 7^{\frac{1}{2}} \).
(iii) We have \( (\sqrt[3]{49})^{5} \). First, write \( \sqrt[3]{49} \) using a fractional index. We know \( 49 = 7^2 \).
So, \( \sqrt[3]{49} = \sqrt[3]{7^2} = (7^2)^{\frac{1}{3}} \).
Now, substitute this back into the original expression:
\( (\sqrt[3]{49})^{5} = ((7^2)^{\frac{1}{3}})^5 \).
Using the exponent rule \( (a^m)^n = a^{m \times n} \) twice:
\( ((7^2)^{\frac{1}{3}})^5 = (7^{2 \times \frac{1}{3}})^5 = (7^{\frac{2}{3}})^5 = 7^{\frac{2}{3} \times 5} = 7^{\frac{10}{3}} \).
(iv) We have \( (\frac{1}{\sqrt[3]{100}})^{7} \). First, write \( \sqrt[3]{100} \) using a fractional index. We know \( 100 = 10^2 \).
So, \( \sqrt[3]{100} = \sqrt[3]{10^2} = (10^2)^{\frac{1}{3}} = 10^{\frac{2}{3}} \).
Now, the expression becomes \( (\frac{1}{10^{\frac{2}{3}}})^{7} \).
Using the exponent rule \( \frac{1}{a^n} = a^{-n} \):
\( (\frac{1}{10^{\frac{2}{3}}})^{7} = (10^{-\frac{2}{3}})^{7} \).
Using the exponent rule \( (a^m)^n = a^{m \times n} \):
\( (10^{-\frac{2}{3}})^{7} = 10^{-\frac{2}{3} \times 7} = 10^{-\frac{14}{3}} \).
In simple words: We are changing roots into powers with fractions. For example, a square root is like raising to the power of one-half. If there's a power outside the root, we multiply the powers together. If a number is in the bottom of a fraction, its power becomes negative.

🎯 Exam Tip: Always simplify the number inside the root to its smallest base possible (e.g., \( 49 = 7^2 \)) before applying fractional index rules. This prevents errors in multiplication.

 

Question 5. Find the 5th root of:
(i) 32
(ii) 243
(iii) 100000
(iv) \( \frac{1024}{3125} \)
Answer:
(i) We need to find the 5th root of 32, which is \( \sqrt[5]{32} \).
First, we find the prime factorization of 32:

So, \( 32 = 2 \times 2 \times 2 \times 2 \times 2 = 2^5 \).
Therefore, \( \sqrt[5]{32} = \sqrt[5]{2^5} = (2^5)^{\frac{1}{5}} = 2^{5 \times \frac{1}{5}} = 2^1 = 2 \).
(ii) We need to find the 5th root of 243, which is \( \sqrt[5]{243} \).
From Question 3 (ii), we know that \( 243 = 3^5 \).
Therefore, \( \sqrt[5]{243} = \sqrt[5]{3^5} = (3^5)^{\frac{1}{5}} = 3^{5 \times \frac{1}{5}} = 3^1 = 3 \).
(iii) We need to find the 5th root of 100000, which is \( \sqrt[5]{100000} \).
We can write \( 100000 = 10 \times 10 \times 10 \times 10 \times 10 = 10^5 \).
Therefore, \( \sqrt[5]{100000} = \sqrt[5]{10^5} = (10^5)^{\frac{1}{5}} = 10^{5 \times \frac{1}{5}} = 10^1 = 10 \).
(iv) We need to find the 5th root of \( \frac{1024}{3125} \), which is \( \sqrt[5]{\frac{1024}{3125}} \).
First, find the prime factorization for 1024 and 3125.

From this, we find \( 1024 = 2^{10} \) and \( 3125 = 5^5 \).
So, \( \sqrt[5]{\frac{1024}{3125}} = \sqrt[5]{\frac{2^{10}}{5^5}} \).
This can be written as \( (\frac{2^{10}}{5^5})^{\frac{1}{5}} \).
Applying the power to both numerator and denominator:
\( \frac{(2^{10})^{\frac{1}{5}}}{(5^5)^{\frac{1}{5}}} = \frac{2^{10 \times \frac{1}{5}}}{5^{5 \times \frac{1}{5}}} = \frac{2^2}{5^1} = \frac{4}{5} \).
In simple words: We need to find a number that, when multiplied by itself five times, gives the original number. For a fraction, we find the 5th root of the top number and the 5th root of the bottom number separately.

🎯 Exam Tip: When finding the nth root, always express the number inside the root as a power of a base. If the base's exponent is a multiple of n, the calculation becomes straightforward.